Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+4y &= 1 \\ -8x+8y &= -6\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-8x = -8y-6$ Divide both sides by $-8$ to isolate $x$ $x = {y + \dfrac{3}{4}}$ Substitute this expression for $x$ in the first equation. $8({y + \dfrac{3}{4}}) + 4y = 1$ $8y + 6 + 4y = 1$ Simplify by combining terms, then solve for $y$ $12y + 6 = 1$ $12y = -5$ $y = -\dfrac{5}{12}$ Substitute $-\dfrac{5}{12}$ for $y$ in the top equation. $8x+4( -\dfrac{5}{12}) = 1$ $8x-\dfrac{5}{3} = 1$ $8x = \dfrac{8}{3}$ $x = \dfrac{1}{3}$ The solution is $\enspace x = \dfrac{1}{3}, \enspace y = -\dfrac{5}{12}$.